Chip select LED using darlington sink

Discussion in 'General Electronics Chat' started by StephenDJ, Oct 27, 2008.

  1. StephenDJ

    Thread Starter Active Member

    May 31, 2008
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    I am trying to use a TTL input darlington sink driver chip to sink an active low LED down away the from the +5 volt rail and thus making it light when the chip select goes low. See circuit .pdf below. While the idea of sinking the cathode down to ground with the anode tied to +5V through a resistor seems easy enough and electrically should work, the only problem I've encountered is that the LED does not have enough time to fully light because the chip select signal only last 1.1 us (one cpu clock cycle). This is a peripheral and I want to see when there's activity on it. The purpose of the capacitor is to let it charge simultaneously while sink is occuring, and then when the sinking stops there will still be a little charge left to make the LED light for a longer period time. My question is, with the right component values, is this possible?
     
  2. Bailey45

    Member

    Oct 27, 2008
    18
    0
    The addition of the capacitor will lengthen the LED on time, however, by the time you have a large enough value you will disrupt the timing you are trying to see. A better solution is to add a 555 timer in one shot mode and drive the LED from its output.
     
  3. StephenDJ

    Thread Starter Active Member

    May 31, 2008
    58
    0
    Yep I thought that's what I need (a 555), but I'd like to explore the idea a little further. As I recall the capacitor is in parallel with the R/LED network, therefore the same voltage from the power supply is applied accross both branches. By "disrupt" I assume you are talking about the capacitor effectively shorting out the R/LED network. But if it's from the power supply, the whole power supply would have to come down before the LED would quit to shine immediately when sink starts. Would not the supply voltage be present and sufficient to power both the LED AND the capacitor immediately when sink starts?
     
  4. Bailey45

    Member

    Oct 27, 2008
    18
    0
    By disrupt I meant the output of the darlington could only be used for the LED as the addition of the capacitor will change the signal if used for any other input. Your power supply should be fine to power this circuit. The length of on time would be roughly R*C. With a low current LED (2mA) R could be 1800Ω and with a 4u7 capacitor the on time would be lengthen by approximatly 8 ms.
     
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